JEE MAIN - Mathematics (2021 - 31st August Morning Shift - No. 17)
A point z moves in the complex plane such that $$\arg \left( {{{z - 2} \over {z + 2}}} \right) = {\pi \over 4}$$, then the minimum value of $${\left| {z - 9\sqrt 2 - 2i} \right|^2}$$ is equal to _______________.
Answer
98
Explanation
Let $$z = x + iy$$
$$\arg \left( {{{x - 2 + iy} \over {x + 2 + iy}}} \right) = {\pi \over 4}$$
$$\arg (x - 2 + iy) - \arg (x + 2 + iy) = {\pi \over 4}$$
$${\tan ^{ - 1}}\left( {{y \over {x - 2}}} \right) - {\tan ^{ - 1}}\left( {{y \over {x + 2}}} \right) = {\pi \over 4}$$
$${{{y \over {x - 2}} - {y \over {x + 2}}} \over {1 + \left( {{y \over {x - 2}}} \right).\left( {{y \over {x + 2}}} \right)}} = \tan {\pi \over 4} = 1$$
$${{xy + 2y - xy + 2y} \over {{x^2} - 4 + {y^2}}} = 1$$
$$4y = {x^2} - 4 + {y^2}$$
$${x^2} + {y^2} - 4y - 4 = 0$$
locus is a circle with center (0, 2) & radius = $$2\sqrt 2 $$
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min. value = $${(AP)^2} = {(OP - OA)^2}$$
$$ = {\left( {9\sqrt 2 - 2\sqrt 2 } \right)^2}$$
$$ = {\left( {7\sqrt 2 } \right)^2} = 98$$
$$\arg \left( {{{x - 2 + iy} \over {x + 2 + iy}}} \right) = {\pi \over 4}$$
$$\arg (x - 2 + iy) - \arg (x + 2 + iy) = {\pi \over 4}$$
$${\tan ^{ - 1}}\left( {{y \over {x - 2}}} \right) - {\tan ^{ - 1}}\left( {{y \over {x + 2}}} \right) = {\pi \over 4}$$
$${{{y \over {x - 2}} - {y \over {x + 2}}} \over {1 + \left( {{y \over {x - 2}}} \right).\left( {{y \over {x + 2}}} \right)}} = \tan {\pi \over 4} = 1$$
$${{xy + 2y - xy + 2y} \over {{x^2} - 4 + {y^2}}} = 1$$
$$4y = {x^2} - 4 + {y^2}$$
$${x^2} + {y^2} - 4y - 4 = 0$$
locus is a circle with center (0, 2) & radius = $$2\sqrt 2 $$
min. value = $${(AP)^2} = {(OP - OA)^2}$$
$$ = {\left( {9\sqrt 2 - 2\sqrt 2 } \right)^2}$$
$$ = {\left( {7\sqrt 2 } \right)^2} = 98$$
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